If the ration of the masses of `SO_(3)` and `O_(2)` gases confined in a vessel is `1 : 1` , then the ratio of their partial pressure would be

Options

A. `5 : 2`
B. `2 : 5`
C. `2 : 1`
D. `1 : 2`

Answer

B

Explanation

`SO_(3) = O_(2) = 10 g :. N_(SO_(3)) = (w)/(80), n_(O_(2)) = (w)/(32)`
The partial pressures will be in the ratio of their moles
Hence `p_(SO_(3)) : p_(SO_(2)) = (w)/(80) : (w)/(32) = (1)/(80) : (1)/(32) = (1)/(5) : (1)/(2) " or " 2 : 5`